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Abhishek Gokhale Class 11 Tuition trainer in Mumbai/>

Abhishek Gokhale

Mathematics and Physics enthusiast

Girgaon, Mumbai, India - 400004.

Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

I am currently in final semester BE M TECH from Mumbai University.I want to teach Maths and/or Physics preferably for Class XI and XII.
I think my basics are clear in Physics and Maths so I have decided to teach them. I have scored 91.4% in Class 10 (CBSE) and 83.3% in Class 12.
I think a strong foundation is necessary for pursuing Science and I would like to help out students to build that.

Languages Spoken

Marathi

English

Hindi

Education

Mumbai University Pursuing

Bachelor of Engineering (B.E.)

Address

Girgaon, Mumbai, India - 400004

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Physics, Mathematics

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Physics, Mathematics

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

State, CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 11 Tuition and Class 12 Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for less than a year.

Answers by Abhishek Gokhale (1)

Answered on 06/01/2016 Learn Tuition/Class XI-XII Tuition (PUC)

Dear Ghufran, First let us assume the variable which we have to find (say X) which represents the quantity of mixture to be drawn off and replaced with water. Now when we draw X amount of mixture it will contain 3 parts of water and 5 parts of syrup (total 8 parts) and also it is replaced by water... ...more
Dear Ghufran, First let us assume the variable which we have to find (say X) which represents the quantity of mixture to be drawn off and replaced with water. Now when we draw X amount of mixture it will contain 3 parts of water and 5 parts of syrup (total 8 parts) and also it is replaced by water so, Parts of water in new mixture = 3 - 3 X/8 + X Parts of syrup in new mixture = 5 - 5X/8 Now we require that the mixture should be half water and half syrup i.e 4 parts of syrup and 4 parts of water so we equate any of the above equations with 4 for eg 3 - 3X/8 +X = 4 which on solving will give X=8/5 Also you can equate the equations to get the answer since we require equal ( half each) parts of syrup and water 3 - 3X/8 + X = 5 -5X/8 Which on solving will also give the same answer X =8/5 Hence 8/5th of the mixture should be drawn off and replaced by water to get half syrup and half water.
Answers 24 Comments
Dislike Bookmark

Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Physics, Mathematics

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Physics, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Physics, Mathematics

No Reviews yet!

Answers by Abhishek Gokhale (1)

Answered on 06/01/2016 Learn Tuition/Class XI-XII Tuition (PUC)

Dear Ghufran, First let us assume the variable which we have to find (say X) which represents the quantity of mixture to be drawn off and replaced with water. Now when we draw X amount of mixture it will contain 3 parts of water and 5 parts of syrup (total 8 parts) and also it is replaced by water... ...more
Dear Ghufran, First let us assume the variable which we have to find (say X) which represents the quantity of mixture to be drawn off and replaced with water. Now when we draw X amount of mixture it will contain 3 parts of water and 5 parts of syrup (total 8 parts) and also it is replaced by water so, Parts of water in new mixture = 3 - 3 X/8 + X Parts of syrup in new mixture = 5 - 5X/8 Now we require that the mixture should be half water and half syrup i.e 4 parts of syrup and 4 parts of water so we equate any of the above equations with 4 for eg 3 - 3X/8 +X = 4 which on solving will give X=8/5 Also you can equate the equations to get the answer since we require equal ( half each) parts of syrup and water 3 - 3X/8 + X = 5 -5X/8 Which on solving will also give the same answer X =8/5 Hence 8/5th of the mixture should be drawn off and replaced by water to get half syrup and half water.
Answers 24 Comments
Dislike Bookmark

Abhishek Gokhale describes himself as Mathematics and Physics enthusiast. He conducts classes in Class 11 Tuition and Class 12 Tuition. Abhishek is located in Girgaon, Mumbai. Abhishek takes at students Home and Online Classes- via online medium. Abhishek is pursuing Bachelor of Engineering (B.E.) from Mumbai University. HeĀ is well versed in Marathi, English and Hindi.

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